Question

当我不知道单词的长度时，如何将单词分开？

Split the string into the specified number of sub segments where each sub
segment consists of a contiguous block of characters. Finally, reverse all the sub segments divided by
separator characters to build a new String and display it.

例如，如果输入为：

String = appleisgreat
ss =4
separator: ,

我想得到结果：

eat,sgr,lei,app

我做到了这一点，我无法将这些词分成特定的子段。

string = input("Enter a stirng:")
ss = eval(input("Enter the number of sub segments:"))
separator = eval(input("Enter the separator:"))

worldlist = list(string)
separate = worldlist/[ss]
print(separate)

Answer 1

您可以计算单词的长度您知道＆＃34;句子＆＃34;的长度，或者起始字符串：len(string)或len(sentence)。
（我避免使用string或String作为var名称，因为它们在许多语言中都被保留为数据类型。）

您知道所需的字词数量为ss（我将调用此wordLength）。

每个单词的长度为len(sentence) // wordLength - 如果它们保证可以被整除。否则，请使用：

wordLength = len(sentence) // wordLength

# // TRUNCATES, so if if its not evenly divisible, 
# the "fractional" number of letters would get left out.
# instead, let's increase all the other the word lengths by one, and now 
# the last word will have the remaining letters.
if len(sentence) % wordLength == 0:
    wordLength += 1

现在查看完整代码：

sentence = "appleisgreat"
ss = 4
seperator = ","

numWords = ss  # rename this variable to be descriptively consistent with my other vars

wordLength = len(sentence) // numWords   # use // to truncate to an integer
print(wordLength)
## 3

# create a list of ss strings, each of length segmentSize
wordlist = []
for wordNum in range(numWords):
  startIndex = wordNum * wordLength
  # print (startIndex, startIndex + wordLength) ## 0 3, 3 6, 6 9, 9 12
  word = sentence[startIndex : startIndex + wordLength]

  # since you want the list in reverse order, add new word to beginning of list.  
  # If reverse order is not required, `append` could be used instead, as wordlist.append(word)
  wordlist.insert(0, word)
print(wordlist)
## ["eat", "sgr", "lei", "app"]    

# lists are iterables, so `join` can be used here to "join" the strings together, seperated by "seperator"
result = seperator.join(wordlist)

print(result)
## "eat,sgr,lei,app"

显然，有更简洁的方法可以完成这项任务。

Answer 2

您只需导入textwrap即可完成此操作。

import textwrap
String ="appleisgreat"
ss=4
print (textwrap.wrap(String, ss-1))

输出：

['app', 'lei', 'sgr', 'eat']

Answer 3

普通蟒蛇：

>>> s = 'appleisgreat'
>>> ss = 4
>>> L = len(s)/ss
>>> separator = ","
>>> separator.join([s[i:i+L] for i in range(0,len(s),L)][::-1])
'eat,sgr,lei,app'

更好，让它成为一个功能：

def make_parts(s, ss, separator):
    # Find out what should be the length of each part
    L = len(s)/ss

    # range(0, len(s), L) is providing desired indices, e.g: 0, 4, 8, etc
    # s[i:i+L] is providing the parts
    # [::-1] is reversing the array
    # str join() method is combining the parts with given separator
    return separator.join([s[i:i+L] for i in range(0,len(s),L)][::-1])

并打电话给

>>> make_parts('appleisgreat', 4, ',')
'eat,sgr,lei,app'

我怎么能把一个不知道有多少的单词分开

3 个答案: